A degree sum condition on the order, the connectivity and the independence number for Hamiltonicity

In [Graphs Combin.~24 (2008) 469--483.], the third author and the fifth author conjectured that if $G$ is a $k$-connected graph such that $\sigma_{k+1}(G) \ge |V(G)|+\kappa(G)+(k-2)(\alpha(G)-1)$, then $G$ contains a Hamiltonian cycle, where $\sigma_{k+1}(G)$, $\kappa(G)$ and $\alpha(G)$ are the minimum degree sum of $k+1$ independent vertices, the connectivity and the independence number of $G$, respectively. In this paper, we settle this conjecture. This is an improvement of the result obtained by Li: If $G$ is a $k$-connected graph such that $\sigma_{k+1}(G) \ge |V(G)|+(k-1)(\alpha(G)-1)$, then $G$ is Hamiltonian. The degree sum condition is best possible.

1 Introduction 1.1 Degree sum condition for graphs with high connectivity to be Hamiltonian In this paper, we consider only finite undirected graphs without loops or multiple edges. For standard graph-theoretic terminology not explained, we refer the reader to [5].
A Hamiltonian cycle of a graph is a cycle containing all the vertices of the graph. A graph having a Hamiltonian cycle is called a Hamiltonian graph. The Hamiltonian problem has long been fundamental in graph theory. Since it is NP-complete, no easily verifiable necessary and sufficient condition seems to exist. Then instead of that, many researchers have investigated sufficient conditions for a graph to be Hamiltonian. In this paper, we deal with a degree sum type condition, which is one of the main stream of this study.
We introduce four invariants, including degree sum, which play important roles for the existence of a Hamiltonian cycle. Let G be a graph. The number of vertices of G is called its order, denoted by n(G). A set X of vertices in G is called an independent set in G if no two vertices of X are adjacent in G. The independence number of G is defined by the maximum cardinality of an independent set in G, denoted by α(G). For two distinct vertices x, y ∈ V (G), the local connectivity κ G (x, y) is defined to be the maximum number of internally-disjoint paths connecting x and y in G. A graph G is k-connected if κ G (x, y) ≥ k for any two distinct vertices x, y ∈ V (G). The connectivity κ(G) of G is the maximum value of k for which G is k-connected. We denote by N G (x) and d G (x) the neighbor and the degree of a vertex x in G, respectively. If α(G) ≥ k, let σ k (G) = min x∈X d G (x) : X is an independent set in G with |X| = k ; otherwise let σ k (G) = +∞. If the graph G is clear from the context, we simply write n, α, κ and σ k instead of n(G), α(G), κ(G) and σ k (G), respectively.
One of the main streams of the study of the Hamiltonian problem is, as mentioned above, to consider degree sum type sufficient conditions for graphs to have a Hamiltonian cycle. We list some of them below. (Each of the conditions is best possible in some sense.) Theorem 1. Let G be a graph of order at least three. If G satisfies one of the following, then G is Hamiltonian.
(i) (Dirac [7]) The minimum degree of G is at least n 2 .
To be exact, Theorem 1 (iii) is not a degree sum type condition, but it is closely related. Bondy [3] showed that Theorem 1 (iii) implies (ii). The current research of this area is based on Theorem 1 (iii). Let us explain how to expand the research from Theorem 1 (iii): Let G be a k-connected graph, and suppose that one wants to consider whether G is Hamiltonian. If α ≤ k, then it follows from Theorem 1 (iii) that G is Hamiltonian. Hence we may assume that α ≥ k + 1, that is, G has an independent set of order k + 1. Thus, it is natural to consider a σ k+1 condition for a k-connected graph. Bondy [4] gave a σ k+1 condition of Theorem 1 (iv).
In this paper, we give a much weaker σ k+1 condition than that of Theorem 1 (iv).
Theorem 2. Let k be an integer with k ≥ 1 and let G be a k-connected graph. If σ k+1 ≥ n + κ + (k − 2)(α − 1), then G is Hamiltonian. Theorem 2 was conjectured by Ozeki and Yamashita [15], and has been proven for small integers k: The case k = 2 of Theorem 2 coincides Theorem 1 (v). The cases k = 1 and k = 3 were shown by Fraisse and Jung [8], and by Ozeki and Yamashita [15], respectively.

Best possibility of Theorem 2
In this section, we show that the σ k+1 condition in Theorem 2 is best possible in some senses.
We first discuss the lower bound of the σ k+1 condition. For an integer l ≥ 2 and l vertex-disjoint graphs H 1 , . . . , H l , we define the graph H 1 + · · · + H l from the union of H 1 , . . . , H l by joining every vertex of H i to every vertex of H i+1 for 1 ≤ i ≤ l − 1. Fix an integer k ≥ 1. Let κ, m and n be integers with k ≤ κ < m and 2m + 1 ≤ n ≤ 3m − κ. Let G 1 = K n−2m + K κ + K m + K m−κ , where K l denotes a complete graph of order l and K l denotes the complement of K l . Then α(G 1 ) = m + 1, κ(G 1 ) = κ and (Note that it follows from condition "n ≤ 3m − κ" that n − 2m − 1 + κ < m.) Since deleting all the vertices in K κ and those in K m−κ breaks G 1 into m + 1 components, we see that G 1 has no Hamiltonian cycle. Therefore, the σ k+1 condition in Theorem 2 is best possible.
We next discuss the relation between the coefficient of κ and that of α − 1. By Theorem 1 (iii), we may assume that α ≥ κ + 1. This implies that for arbitrarily ε > 0. Then one may expect that the σ k+1 condition in Theorem 2 can be replaced with "n + (1 + ε)κ + (k − 2 − ε)(α − 1)" for some ε > 0. However, the graph G 1 as defined above shows that it is not true: For any ε > 0, there exist two integers m and κ such that ε(m − κ) ≥ 1. If we construct the above graph G 1 from such integers m and κ, then we have but G 1 is not Hamiltonian. This means that the coefficient 1 of κ and the coefficient k − 2 of α − 1 are, in a sense, best possible.

Comparing Theorem 2 to other results
In this section, we compare Theorem 2 to Theorem 1 (iv) and Ota's result (Theorem 3).
We next compare Theorem 2 to the following Ota's result.

Notation and lemmas
Let G be a graph and H be a subgraph of G, and let x ∈ V (G) and X ⊆ V (G). We denote by N G (X) the set of vertices in V (G)\X which are adjacent to some vertex in If there is no fear of confusion, we often identify H with its vertex set V (H). For example, we often write G − H instead of G − V (H). For a subgraph H, a path P is called an H-path if both end vertices of P are contained in H and all internal vertices are not contained in H. Note that each edge of H is an H-path.
Let C be a cycle (or a path) with a fixed orientation in a graph G. For x, y ∈ V (C), we denote by C[x, y] the path from x to y along the orientation of C. The reverse sequence of C[x, y] is denoted by . We denote C[x, y]−{x, y}, C[x, y]− {x} and C[x, y] − {y} by C(x, y), C(x, y] and C[x, y), respectively. For x ∈ V (C), we denote the successor and the predecessor of x on C by x + and x − , respectively. For X ⊆ V (C), we define X + = {x + : x ∈ X} and X − = {x − : x ∈ X}. Throughout this paper, we consider that every cycle has a fixed orientation.
In this paper, we extend the concept of insertible, introduced by Ainouche [1], which has been used for the proofs of the results on cycles.
Let G be a graph, and H be a subgraph of G.
Lemma 1. Let D be a cycle of a graph G. Let k be a positive integer and let Q 1 , Q 2 , . . . , Q k be paths of G −D with fixed orientations such that V (Q i ) ∩V (Q j ) = ∅ for 1 ≤ i < j ≤ k. If the following (I) and (II) hold, then G[V (D ∪Q 1 ∪Q 2 ∪· · ·∪Q k )] is Hamiltonian.
Proof. We can easily see that is such a cycle. By the choice of u 1 and v 1 , . Hence by repeating this argument, we can obtain a cycle D * so that |C| is as large as possible. Now, we change the "base" cycle from D to C, and use the symbol (·) + for the orientation of C. Suppose that V (Q i − C) = ∅ for some i with i ∈ {1, 2, . . . , k}. We may assume In the former case, let C ′ = wC[z + , z]w, and in the latter case, let In the rest of this section, we fixed the following notation. Let C be a longest cycle in a graph G, and H 0 be a component of is Hamiltonian, which contradicts the maximality of C.
Then the following hold (see Figure 1).
Proof. Let P 0 be a C-path which connects u 1 and u 2 , and V (P 0 ) ∩ V (H 0 ) = ∅. We first show (i) and (ii). Suppose that the following (a) or (b) holds for some v 1 ∈ C(u 1 , x 1 ] and some v 2 ∈ C(u 2 , x 2 ]: (a) There exists a C-path P 1 joining v 1 and v 2 . (b) There exist disjoint C-paths P 2 joining v l and w, and P 3 joining v 3−l and w − for some l ∈ {1, 2} and some w ∈ C(v l , u 3−l ]. We choose such vertices v 1 and v 2 so that |C[u 1 , v 1 ]| + |C[u 2 , v 2 ]| is as small as possible. Without loss of generality, we may assume that By Lemma 2, we can obtain the following statement (1), and by the choice of v 1 and v 2 , we can obtain the following statements (2)-(5): (4) I(x; C) ∩ I(y; C) = ∅ for x ∈ V (Q 1 ) and y ∈ V (Q 2 ).
By using similar argument as above, we can also show (iii) and (iv). We only prove (iii). Suppose that for some v 1 ∈ C(u 1 , . By the choice of v 1 and v 2 and Lemma 3 (ii), . Hence by applying Lemma 1 as Q 1 and Q 2 , we see that there exits a longer cycle than C, a contradiction.

Proof of Theorem 2
Proof of Theorem 2. The cases k = 1, k = 2 and k = 3 were shown by Fraisse and Jung [8], by Bauer et al. [2] and by Ozeki and Yamashita [15], respectively. Therefore, we may assume that k ≥ 4. Let G be a graph satisfying the assumption of Theorem 2. By Theorem 1 (iii), we may assume α(G) ≥ κ(G) + 1. Let C be a longest cycle in G. If C is a Hamiltonian cycle of G, then there is nothing to prove. Hence we may assume that G − V (C) = ∅. Let H = G − V (C) and x 0 ∈ V (H). Choose a longest cycle C and x 0 so that d C (x 0 ) is as large as possible.
Note that m ≥ κ(G) ≥ k. Let Let u ′ i be the vertex in N C (H 0 ) such that C(u i , u ′ i ) ∩ N C (H 0 ) = ∅. By Lemma 2, there exists a non-insertible vertex in C(u i , u ′ i ). Let x i ∈ C(u i , u ′ i ) be the first non-insertible vertex along the orientation of C for each i ∈ M 1 , and let We check the degree of x i in C and H. Since x i is non-insertible, we can see that By the definition of x i , we clearly have and We check the degree sum in C of two vertices in X. Let i and j be distinct two integers in M 1 . In this paragraph, we let By Lemma 3 (i) and since N H 0 (x i ) = ∅ for i ∈ M 1 , we obtain the following.
Proof. Let s and t be distinct two integers in M 1 . By the inequality (4), we have Let I be a subset of M 0 such that |I| = k + 1 and {0, s, t} ⊆ I. By Claim 1, {x i : i ∈ I} is an independent set. By the inequality (1), we deduce i∈I\{0,s,t} By the inequality (2) and the definition of I, we obtain Thus, it follows from these three inequalities that Let S be a cut set with |S| = κ(G), and let V 1 , V 2 , . . . , V p be the components of G − S. By Claim 2, we may assume that there exists an integer l such that C[u l , u ′ l ) ⊆ V 1 . By Lemma 3 (i), we obtain By replacing the labels x 2 and x 3 if necessary, we may assume that x 1 , x 2 and x 3 appear in this order along the orientation of C. In this paragraph, the indices are taken modulo 3. From now we let for each i ∈ {1, 2, 3}, and let W := W 1 ∪ W 2 ∪ W 3 (see Figure 2). Note that   Proof. We first show that D ∪ X ∪ W ⊆ V 1 ∪ S. Suppose not. Without loss of generality, we may assume that there exists an integer h in Since v ∈ V 2 , it follows from Lemma 3 (i) and (ii) that Let I be a subset of M 0 \ {h} such that |I| = k and {0, l} ⊆ I. By Claim 1 and Lemma 3 (i) and (ii), {x i : i ∈ I} ∪ {v} is an independent set of order k + 1. By the above inequality and the inequality (5), we obtain On the other hand, the inequality (1) yields that i∈I\{0,l} By the above two inequalities, we deduce Combining this inequality with the above inequality, we get i∈I d G ( We next show that H − H 0 ⊆ V 1 ∪ S. Suppose not. Without loss of generality, we may assume that there exists a vertex y ∈ Hence, by the same argument as above, we can obtain a contradiction. Thus we may assume that N C (H y ) ∩ (D i ∪ {x i }) = ∅ for all i ∈ M 1 \ {l}. Then, since y ∈ V 2 and Let I be a subset of M 0 such that |I| = k and {0, l} ⊆ I. Since x l ∈ V 1 , y ∈ V 2 , H y = H 0 and N C (H y ) ∩ (D i ∪ {x i }) = ∅ for all i ∈ M 1 \ {l}, it follows from Claim 1 that {x i : i ∈ I} ∪ {y} is an independent set of order k + 1. By the above inequality and the inequality (5), we obtain Therefore, by the above inequality and the inequality (1), we obtain Combining the above two inequalities, We finally show that H 0 ⊆ V 1 ∪ S. Suppose not. Without loss of generality, we may assume that there exists a vertex y 0 ∈ H 0 ∩ V 2 . Then Since u l ∈ V 1 , we have H 0 ∩ S = ∅. Note that by the above argument, X ⊆ V 1 ∪ S. Therefore, by Claim 2, Therefore, we can improve the inequality (4) as follows: By the inequality (1) and the inequality (3), Hence, by the above four inequalities, we deduce d G (y 0 ) By Claim 3, there exists an integer r such that Choose r and v 2 so that v 2 = u ′ r if possible. Without loss of generality, we may assume that v 2 ∈ V 2 . Note that Proof. Let w ∈ W . Without loss of generality, we may assume that w ∈ W 1 . Then by applying Lemma 1 as Q 1 = D 1 , Q 2 = D 2 and where P [u 2 , u 1 ] is a C-path passing through some vertex of H 0 , we can obtain a cycle C ′ such that V (C) \ {w} ⊆ V (C ′ ) and V (C ′ ) ∩ V (H 0 ) = ∅ (note that (I) and (II) of Lemma 1 hold, by Lemma 3 (i) and (ii) and the definition of insertible and D i ). Note that by the maximality of |C|, |C ′ | = |C|. Note also that d C ′ (w) ≥ d C (w). By the choice of C and x 0 , we have d C ′ (w) ≤ d C (x 0 ), and hence by Claim 1 and the fact that By Lemma 3 and Claim 3, we have Moreover, by Lemma 3 and Claim 1, the following claim holds.
We now check the degree sum of the vertices x 1 , x 2 and x 3 in C. In this paragraph, the indices are taken modulo 3. By Lemma 3 (ii), and let L = i∈{1,2,3} L i (see Figure 3).
In this proof, we assume x l = x 1 (recall that l is an integer such that C[u l , u ′ l ) ⊆ V 1 , see the paragraph below the proof of Claim 2). We divide the proof into two cases.
Proof. Suppose that |W | ≥ κ(G) + k − 4. By Claim 3, we obtain Let W ′ be a subset of (W ∪ {x 0 , x 1 , x 2 , x 3 }) ∩ V 1 such that |W ′ | = k and x 1 ∈ W ′ . Since W ′ ⊆ V 1 and v 2 ∈ V 2 , it follows from Claim 5 that W ′ ∪ {v 2 } is an independent set of order k + 1. By the inequality (5) and Claims 3 and 4, we obtain . By the inequality (1) and Claim 4, By the above two inequalities, we obtain Therefore, since w∈W ′ d H (w) ≤ |H ∩ (V 1 ∪ S)| − 1 by the inequality (7), it follows that Summing this inequality and the inequality (6) By the assumption of Case 1, we can take a subset W * of W ∪ {x 0 } such that |W * | = k − 2. By Claim 5, W * ∪ {x 1 , x 2 , x 3 } is independent. Moreover, by Claim 4 and the assumption that By Subclaim 7.1, summing this inequality and the inequality (9) yields that Case 2. |W | ≤ k − 4. By Claim 6, we can take a subset L * of L such that |L * | = k − 3 − |W |. Let I = {i : x i ∈ L * }. By Claim 5, W ∪ L * ∪ {x 0 , x 1 , x 2 , x 3 } is an independent set of order k + 1. By the inequality (8), we have On the other hand, it follows from Claim 4, the assumption d C (x 0 ) ≤ α − 2 and the inequality (1) that By the inequality (7), we obtain Summing the above two inequalities yields that

By Cases 1 and 2, we have
This completes the proof of Claim 7.
Claim 9. If there exist distinct two integers s and t in M 1 such that u s ∈ N C (x t ), Proof. Suppose that there exists a vertex z ∈ N C (x s )∩C[u t , u s ] such that z ∈ U. We show that X ∪ {x 0 , z + } is an independent set of order |X| + 2. By Claim 5, we only show that z + ∈ X and z + ∈ N C (x i ) for each x i ∈ X ∪ {x 0 }. Since z ∈ U, it follows from Lemma 3 (i) that z + ∈ X. Suppose that z + ∈ N C (x h ) for some x h ∈ X ∪ {x 0 }. Since x s is a non-insertible vertex, it follows that x h = x s . Let z s be the vertex in C(u s , x s ] such that z ∈ N G (z s ) and z ∈ N G (v) for all v ∈ C(u s , z s ). By Lemma 3 (ii), we obtain x h ∈ C[u ′ s , z]. Therefore, x h ∈ C(z, u s ] ∪ {x 0 }. If x h ∈ C(z, u s ], then we let z h be the vertex in C(u h , x h ] such that z + ∈ N G (z h ) and z + ∈ N G (v) for all v ∈ C(u h , z h ). We define the cycle C * as follows (see Figure 4): Then, by similar argument in the proof of Lemma 3, we can obtain a longer cycle than C by inserting all vertices of V (C \ C * ) into C * . This contradicts that C is longest. Hence z + ∈ N C (x h ) for each x h ∈ X ∪ {x 0 }. Thus, by Claim 7, X ∪ {x 0 , z + } is an independent set of order |X| + 2 = α(G) + 1, a contradiction. We divide the rest of the proof into two cases.
Proof. Suppose that |Y | ≤ γ +2. By the assumption of Case 1, we have x 0 v 2 ∈ E(G). Since |M 0 | = |X| + 1 ≥ k + γ + 2 and |Y | ≤ γ + 2, there exists a subset I of M 0 \ {i : x i ∈ Y } such that |I| = k and {0, l} ⊆ I. Then {x i : i ∈ I} ∪ {v 2 } is an independent set of order k + 1. By the inequality (5) and Claims 3 and 7, we obtain Therefore it follows from the inequality (1) that By the inequality (7), i∈I d H (x i ) ≤ |H ∩ (V 1 ∪ S)| − 1. Summing these two inequalities and the inequality (6) yields that Recall that r is an integer such that v 2 ∈ C(x r , u ′ r ] ∩ V 2 (see the paragraph below the proof of Claim 3). In the rest of Case 1, we assume that l = 1. If u ′ r = u 1 , then let r = 2 and u 3 = u ′ 2 ; otherwise, let r = 3 and let u 2 be the vertex with u ′ 2 = u 3 . By Claim 8, we have W ⊆ X. Hence we obtain Y ∪ W ∪ L ⊆ X \ {x 1 }. Recall that W ∩ L = ∅. Therefore, by Claims 6 and 10, we obtain and u h ∈ C(x 1 , u 2 ) holds (especially, if r = 3 then u h ∈ N C (x 2 ) and u h ∈ C(x 1 , u 2 ) holds) (see Figure 5). If r = 2 and u h ∈ N C (x 1 ), then v 2 ∈ C[u 1 , u h ] (see Figure 5 (i)). If r = 2 and u h ∈ N C (x 2 ), then v 2 ∈ C[u 2 , u h ] (see Figure 5 (ii)). If r = 3, then u h ∈ N C (x 2 ) and v 2 ∈ C[u 2 , u h ] (see Figure 5 (iii)). In each case, we obtain a contradiction to Claim 9.
We rename x i ∈ X for i ≥ 1 as follows (see Figure 6): Rename an arbitrary vertex of X as x 1 . For i ≥ 1, we rename x i+1 ∈ X so that u i+1 ∈ N C (x i ) ∩ (U \ {u i }) and |C[u i+1 , x i )| is as small as possible. (For x i ∈ X, let x ′ i and x ′′ i be the successors of x i and x ′ i in X along the orientation of C, respectively. Then by applying Claim 6 as By the definition of x ′ i , x ′′ i and Claim 8, we have W 1 = W 2 = ∅ (note that W ∩ {x 1 , x 2 , x 3 } = ∅). By the definitions of x ′ i , x ′′ i , L 1 and L 2 , we also have L 1 = L 2 = ∅. Thus W 3 ∪ L 3 = ∅. By Lemma 3 (i) and since W ∪ L ⊆ X, this implies that N C (x i ) ∩ (U \ {u i }) = ∅.) Let h be the minimum integer such that x h+1 ∈ C(x h , x 1 ]. Note that this choice implies h ≥ 2. We rename h vertices in X as {x 1 , x 2 , . . . , x h } as above, and m − h vertices in X \ {x 1 , x 2 , . . . , x h } as {x h+1 , x h+2 , . . . , x m } arbitrarily. Let If possible, choose x 1 so that A 2 ∩ U 1 = ∅. We divide the proof of Case 2 according to whether h ≤ k or h ≥ k + 1.
By the choice of {x 1 , . . . , x h }, we have By Claim 9 and (10), we obtain By Lemma 3 (i) and (ii),
By Claims 3 and 7, the assumption of Case 2 and the choice of r and v 2 , we have p i=2 V i ⊆ U = N C (x 0 ). Since x 0 ∈ V 1 ∪ S by Claim 3, this implies that x 0 ∈ S. Claim 11. |X ∩ V 1 | ≤ k − 1.
Recall U 1 = {u i ∈ U : x i ∈ X ∩ V 1 }. By Claim 11, we have |U 1 | ≤ k − 1. By the assumption of Case 2.2 and the choice of x 1 , we obtain A 2 ∩ U 1 = ∅, and hence we can take a subset I of {2, 3, . . . , h} such that |I| = k and {i : A i+1 ∩ U 1 = ∅} ⊆ I. Let X I = {x i : i ∈ I}.
By Claim 5, X I ∪ {x 0 } is an independent set of order k + 1. Let Then, since |C[u i , u ′ i )| ≥ 2 for i ∈ M 1 \ I, the following inequality holds: If x i ∈ X I ∩ S, then it follows from Lemma 3 (i) and Claim 9 that If x i ∈ X I ∩ V 1 , then, by Lemma 3 (i) and Claim 9, Since U ∩ V 2 = ∅, we obtain |U ∩ V 2 | − |B i+1 ∩ U 1 ∩ V 2 | ≥ 1 for all i ∈ I except for at most one, and hence i∈I : By the choice of I, we have On the other hand, since x 0 ∈ S, it follows from Claim 3 that Thus, we deduce By the inequality (2), i∈I∪{0} d H (x i ) ≤ |H| − 1. Hence i∈I∪{0} d G (x i ) ≤ |G| + κ(G) + (k − 2)(α(G) − 1) − 1, a contradiction.